A Piece of P.I.E

As the title suggests, we use PIE, or the Principle of Inclusion-Exclusion. We count the complement: the number of ways at least one corner can have a $3\times 3$ white corner. The number of ways one corner can have a $3\times 3$ white square is $2^7$ , as each of the other 7 squares can be either color. The number of ways two adjacent corners can have white $3\times 3$ squares is $2^4$ , as there are 4 squares that can be either color. The number of ways two opposite corners can have white $3\times 3$ squares is $2^2$ , since only 2 squares can be either color. The number of ways three corners can have white $3\times 3$ squares is $2$ , as only the other corner can be either color. Finally, there is only $1$ way all four corners can have $3\times 3$ white squares.

By PIE, the number of ways is $4\cdot 2^7-4\cdot 2^4-2\cdot 2^2+4\cdot 2-1=447$ Then, $n$ is $2^{16}-447=65089$ and the answer is $6+5+0+8+9=\boxed{28}$ .

C++, since this problem has annoying casework

#include <iostream>
using namespace std;

int main() {
    int ans=0;
    for(int i = 0; i < 65536; i++)
    {
        if((i&1911)&&(i&(1911*2))&&(i&(1911*16))&&(i&(1911*32)))
        {
            ans++;
        }
    }

    cout << ans;
}

Returns 65089.

This is an interesting exercise in bitmasks; can you guess why 1911 was chosen?

((1+2+4)*(1+16+256).)